Given that $\phi$ and $\phi_n,n \geq 1$ are continuous functions from $\mathbb{R}$ into $\mathbb{C}$ such that $ \phi(0) = \phi_n(0) = 1$, $\phi(x) \neq 0$ and $\phi_n(x) \neq 0$ for any $x$. Suppose that for any compact subset $K$ of $\mathbb{R}$ $$ \lim_{n\to \infty} \sup_{x \in K} \vert \phi(x) - \phi_n(x) \vert = 0. $$ Does this imply that $$ \forall x \in \mathbb{R}: \lim_{n \to \infty} \log \phi_n(x) = \log \phi(x)$$?

  • $\begingroup$ No, because the $\log$ function cannot be continuous on $\mathbb C \setminus \{0\}$ -- in particular you can take $\phi, \phi_n$ to be constant mappings. But for someone to give you a counterexample you will need to specify "which" $\log$ on $\mathbb C \setminus \{0\}$ you mean. $\endgroup$ – Mees de Vries Sep 5 '18 at 12:35
  • $\begingroup$ Err, forget the point about constant mappings. That's not true with the condition that $\phi(0) = \phi_n(0) = 1$. You can use a locally constant mapping with value at the a discontinuity of $\log$. $\endgroup$ – Mees de Vries Sep 5 '18 at 12:41
  • $\begingroup$ I choose to define $\log(z) = \log(|z|) + i arg(z) with -\pi < arg(z) \leq \pi $\endgroup$ – M. Ost Sep 6 '18 at 0:43

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